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Fractions

Hi,

I have some queries.

There are some questions on the attachment below of which I am not too sure about.

Indices.JPG

For i)0
ii)9
iii) I'm not too sure

Help would be appreciated.

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Reply 1
i) That's incorrect. Any number to the power of 0 is equal to 1.
ii) Correct.
iii) Use the index law:

(ab)c=ab×c(a^b)^c = a^{b\times c}
Reply 2
Original post by notnek
i) That's incorrect. Any number to the power of 0 is equal to 1.
ii) Correct.
iii) Use the index law:

(ab)c=ab×c(a^b)^c = a^{b\times c}


So would it be -6/4
Reply 3
Original post by zed963
So would it be -6/4


Two negatives make a plus (it should be +6/4 in the exponent).
Reply 4
Original post by f1mad
Two negatives make a plus (it should be +6/4 in the exponent).


So whats 16^6/4
Reply 5
Original post by zed963
So whats 16^6/4


Remember, 16=24 16=2^4
Reply 6
Original post by zed963
So whats 16^6/4


16^(6/4)= 16^(3/2)
Reply 7
Original post by raheem94
Remember, 16=24 16=2^4


I don't understand what u mean here.
Reply 8
Original post by zed963
I don't understand what u mean here.


Raise both sides to the power of 6/4.
Reply 9
Original post by zed963
I don't understand what u mean here.


1664=(24)64 16^{\frac64} = (2^4)^{\frac64}
Reply 10
Original post by raheem94
1664=(24)64 16^{\frac64} = (2^4)^{\frac64}


I would like you to explain why you are doing the things.
Reply 11
would the answer be 18 by any chance?

I meant 2 to the power 6
(edited 11 years ago)
Reply 12
Original post by zed963
I would like you to explain why you are doing the things.


We have got 1664 16^{\frac64}

Now we need to simplify it further, so we will notice that, 16=24 16=2^4

Hence, 1664=(24)64 16^{\frac64} = (2^4)^{\frac64}

Now use the rule, (ax)bd=ax×bd (a^x)^{\frac{b}{d}} = a^{x \times \frac{b}{d}}

1664=(24)64=24×64=26 16^{\frac64} = (2^4)^{\frac64} = 2^{4 \times \frac64} = 2^6

Do you get it?
Reply 13
Original post by zed963
would the answer be 18 by any chance?


The answer is 64.
Reply 14
Original post by zed963
I would like you to explain why you are doing the things.

You seem to be struggling so I'll show you an alternative method that you may be more used to:

1664=1632\displaystyle 16^{\frac{6}{4}} = 16^{\frac{3}{2}}

Raising a number to the power of 32\frac{3}{2} is the same as square-rooting it (the 2 on the denominator tells you this) and then cubing it (from the numerator). So you get:

1632=(16)3=43=64\displaystyle 16^{\frac{3}{2}} = (\sqrt{16})^3 = 4^3 = 64
(edited 11 years ago)
Reply 15
Original post by notnek
You seem to be struggling so I'll show you an alternative method that you may be more used to:

1664=1632\displaystyle 16^{\frac{6}{4}} = 16^{\frac{3}{2}}

Raising a number to the power of 32\frac{3}{2} is the same as squaring it (the 2 on the denominator tells you this) and then cubing it (from the numerator). So you get:

1632=(16)3=43=64\displaystyle 16^{\frac{3}{2}} = (\sqrt{16})^3 = 4^3 = 64


Its just that we've been taught using a different method.
Reply 16
Original post by zed963
Its just that we've been taught using a different method.

Do you understand the method I gave? Which method have you been taught?
Reply 17
Original post by notnek
Do you understand the method I gave? Which method have you been taught?


I've been taught that you have 4 boxes and all those 4 boxes multiplied by something that has to = 16 and now since its asking for 6 we would just multiply the number inside the box to the power of 6.

4 boxes come from the denominator of 6/4
Original post by zed963
I've been taught that you have 4 boxes and all those 4 boxes multiplied by something that has to = 16 and now since its asking for 6 we would just multiply the number inside the box to the power of 6.

4 boxes come from the denominator of 6/4


This seems very odd. Show your method and perhaps we can make sense of it.
Reply 19
Original post by steve2005
This seems very odd. Show your method and perhaps we can make sense of it.


I also can't understand what he is saying. But i also don't see any point for explaining things in this way, a GCSE student shouldn't have any problem doing these in the way it is done in this thread.

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